Adding parametricplot3d in only z axis - wolfram-mathematica

I am trying to add this parametric plot only in the z-axis (right now when I add it expands in the x,y, and z), the effect of this summation would be addition of amplitudes of the sine waves. Here is what I have now. http://imgur.com/j9hN7VR
Here is the code I am using to implement it:
frequency = 1000;
speed = 13397.2441;
wavelength = speed/frequency;
s = (r - 2);
t = (r - 4);
u = (r - 6);
v = (r - 8);
ParametricPlot3D[{{r*Cos[q] - 4, r*Sin[q], Sin[(2*Pi)/wavelength*(r)]},
{s*Cos[q] - 2, s*Sin[q], Sin[(2*Pi)/wavelength*(s + wavelength/4 - 1)]},
{t*Cos[q], t*Sin[q], Sin[(2*Pi)/wavelength*(t + wavelength/4 + 0.5)]},
{u*Cos[q] + 2, u*Sin[q], Sin[(2*Pi)/wavelength*(u + wavelength/4 + 1.65)]},
{v*Cos[q] + 4, v*Sin[q], Sin[(2*Pi)/wavelength*(v + wavelength/4 + 3.5)]}},
{r, 0, 25}, {q, 0, Pi}, PlotPoints -> 30, Mesh -> 20, PlotRange -> {{-25, 25}, {0, 35}, {-6, 6}}]
Any suggestions would be greatly appreciated!

Unfortunately I could not find an answer for this, so I ended up just simulating in MATLAB instead by generating all values over the field (in a matrix) and then summing as I was trying to do here.

Related

Why does not Epilog work in my function in Mathematica?

So in order to save some time, I wrote a function to plot a graph with a lot of default settings. I want to add a 0 tick to the Axes, so I added Epilog in the plot. However, the 0 does not seem to show up in the graph, and the Epilog does not seem to be working at all.
LatexTextStyle[text_] :=
Style[text, FontSize -> 18, FontFamily -> "CMU Serif"]
StyledText[text_] := Text[LatexTextStyle[text]]
StyledTextPos[text_, posx_, posy_] :=
Text[LatexTextStyle[text], {posx, posy}]
NPlot[fns_, variable_ : x, xmin_, xmax_, ymin_, ymax_,
pltStyle_ : Default, epilog_ : {}, marginScale_ : 0.12,
fontSize_ : 18, xOffset_ : 10, yOffset_ : 10] := Plot[
fns,
{variable, xmin, xmax},
PlotStyle -> pltStyle,
AspectRatio -> Equal,
PlotRange -> {
{xmin - marginScale*Abs[xmax - xmin],
xmax + marginScale*Abs[xmax - xmin]}, {ymin -
marginScale*Abs[ymax - ymin],
ymax + marginScale*Abs[ymax - ymin]}
},
BaseStyle -> {FontFamily -> "CMU Serif", FontSize -> fontSize},
AxesStyle -> Arrowheads[{0.0, 0.05}],
Epilog -> {
StyledTextPos["0", -xOffset, -yOffset]
}
]
NPlot[fns = x^2 + 2 x + 1, xmin = -10, xmax = 10, ymin = -5,
ymax = 15]
Here is the graph that I got:
When I tried other Epilog inputs, nothing appeared to be showing up.
Your offset defaults are positioning the 0 at {-10, -10}, below the vertical plot range.
These defaults position the 0 correctly:
NPlot[ ... , xOffset_ : 0.6, yOffset_ : 1] := etc.

3d collision formula based on xyz

Here is the problem. We have two points (spheres) in xyz, with this info:
1- x,y,z => The center of the object is currently located at
2- r => The collision radius of the object
3- Vx, Vy, Vz => object is traveling along the vector. If that vector is (0,0,0), the object is stationary.
Note: The radii and positions are in meters and velocities are measured in meters per second.
Question: For each test, output a single line containing the time (in seconds) since the start of the test at which the two objects will first collide. If they never collide, print No collision instead.
I want to know about the formula of calculation this time. Any idea would be appreciated.
Examples:
1-
xyz(1): -7 5 0
v(1): -1 0 3
r(1): 3
xyz(2): 10 7 -6
v(2): -2 0 4
r(2): 6
t: 8.628 // this is the answer
2-
xyz(1): 10 3 -10
v(1): -9 3 -8
r(1): 5
xyz(2): 2 0 0
v(2): 6
r(2): -4 3 -10
t: 0.492 // this is the answer
To simplify problem, let us use Halileo's principle. Consider the first object as base point, so the second objects moves relative to it.
Put the first object position in coordinate origin.
Subtract the first object initial coordinates from the second one coordinates, do the same for velocity components
x2_0 = x2_0 - x1_0 (same for y,z)
Vx2 = Vx2 - Vx1 (same for y,z)
Now we have second center coordinates against time
x = x2_0 + Vx2 * t
y = y2_0 + Vy2 * t
z = z2_0 + Vz2 * t
and squared distance to origin:
dd = x*x + y*y + z*z =
(x2_0 + Vx2 * t)^2 + ... =
x2_0^2 + 2*x2_0*Vx2*t + Vx2^2*t^2 + ...
and we need to calculate when dd becomes equal to squared radius sum (r1+r2)^2
t^2 * (Vx2^2+Vy2^2+Vz2^2) + t*(2*x2_0*Vx2+2*y2_0*Vy2+2*z2_0*Vz2) +
x2_0^2 + y2_0^2 + y2_0^2 - (r1+r2)^2 = 0
Solve this quadratic equation for t, get 0,1 or 2 solutions.
Case of 0 solutions - no collision
Case of 1 solution with positive t - moment of touching
Case of two solutions - get smaller positive t for the moment of collision.
Negative values of t mean collision "in the past", before the start of the test
Quick test in Python (ideone)
from math import sqrt, isclose
def collisiontime(x1,y1,z1,vx1,vy1,vz1,r1, x2,y2,z2,vx2,vy2,vz2,r2):
x2 -= x1
y2 -= y1
z2 -= z1
vx2 -= vx1
vy2 -= vy1
vz2 -= vz1
a = vx2**2 + vy2**2 + vz2**2
b = 2*x2*vx2 + 2*y2*vy2 + 2*z2*vz2
c = x2**2 + y2**2 + z2**2 - (r1+r2)**2
D = b**2-4*a*c
if D < 0:
return None
if isclose(D, 0):
return -b/2/a
return (-b - sqrt(D)) / 2 /a, (-b + sqrt(D)) / 2 /a
print(collisiontime(0, 0, 0, 2, 0, 0, 2, 25, 0, 0, -3, 0, 0, 3)) # 1=> <=2
print(collisiontime(0, 0, 0, 2, 0, 0, 2, 25, 5, 0, 1, 0, 0, 3)) # 1==> 2=> chase with touching
print(collisiontime(-7, 5, 0,-1, 0, 3, 3, 10, 7, -6, -2, 0, 4, 6))
print(collisiontime(10, 3, -10,-9, 3, -8,5, 2, 0, 0, -4, 3, -10, 6))
(4.0, 6.0)
25.0
(8.627718676730986, 14.372281323269014)
(0.4917797757201004, 3.646151258762658)

slicing 3D plot and dynamic number of controls MATHEMATICA

I am going to have a n-dimensional function and I would like to able plot 2D planes for the rest of the variables fixed to a number. For example:
u = w^2 y^3 + x^5 z^4 + I z^6 + I z^2 Sin[y + x - 1]+k*Sin[w*pi]
I have 5 variables in here and lets assume I want to fix z w plane and plot with sliding y,z, and k. I have many issues to fix to get what I want to do,
1- As it is the code is not working. I need to figure out to update the limsL and limsR for the sliders. If I remove the doloop and the limits at least I dont get the RED error bar. I need to update those values maybe using a button to get all the data and second button to plot.
2- But even with default limits for sliders [0,1] I do not get a plot. The values are updated at the interface. Under varying variables but does not effect the u function. And actually I prefer variables stay as z,y,x etc and not to get numerical values.
Manipulate[DynamicModule[{u =
z Sin[\[Pi] x] + Cos[\[Pi] y] + y^6 Cos[2 \[Pi] y], vars = {x, y, z}, varlims = {{1, 2}, {3, 4}, {5, 6}}, poi = {x, y},
svars = {z, r}, data = Table[RandomReal[], {20}]}, Column[{Style["Ploter", "Function"],
Row[{"Function ", InputField[u]}, Spacer[20]],
Row[{"Variables ", InputField[Dynamic[vars]]}],
Row[{"Variable limits ", InputField[Dynamic[varlims]]}],
Row[{"Plane of interest", InputField[Dynamic[poi]]}],
Row[{"Varying variables", InputField[Dynamic[svars]]}],
Plotslices[u, vars, varlims, poi, svars, size],
Dynamic[
countersvar = Dimensions[svars][[1]];
limsL = ConstantArray[0, countersvar];
limsR = ConstantArray[0, countersvar];
Do[
v = svars[[i]];
posv = Position[vars, v][[1]];
lv = varlims[[posv, 1]][[1]];
rv = varlims[[posv, 2]][[1]];
limsL[[i]] = lv;
limsR[[i]] = rv;
, {i, countersvar}];
Grid[
Table[With[{i = i}, {svars[[i]],
Slider[Dynamic[svars[[i]], {limsL[[1]], limsR[[i]]}]],
Dynamic[svars[[i]]]}], {i, Dimensions[svars][[1]]}]]]
}]], {size, {Small, Medium, Full}}, ControlPlacement -> Bottom,ContinuousAction -> False, Initialization :> (
Plotslices[u_, vars_, varlims_, poi_, svars_, size_] :=
Module[{v1, v2, lv1, lv2, rv1, rv2, posv1, posv2},
v1 = poi[[1]];
v2 = poi[[2]];
posv1 = Position[vars, v1][[1]];
posv2 = Position[vars, v2][[1]];
lv1 = varlims[[posv1, 1]][[1]];
lv2 = varlims[[posv2, 1]][[1]];
rv1 = varlims[[posv1, 2]][[1]];
rv2 = varlims[[posv2, 2]][[1]];
psl =
Plot3D[u, {v1, lv1, rv1}, {v2, lv2, rv2},
PerformanceGoal -> "Quality", Mesh -> None,
ColorFunction -> Function[{v1, v2, z}, Hue[z]],
ImageSize -> size];
Return[psl];];
)
]
I am sorry for the formatting . I tried to put it together crtl+K but it did not work.

Plotting Interpolations in Mathematica

It's apparently very simple but I can't find my mistake. The Plot gives me no points at all.
tmax = 1.;
nmax = 10;
deltat = tmax/nmax;
h[t_, s_] := t^2 + s^2;
T = Table[{{n*deltat}, {n*deltat}, h[n*deltat, n*deltat]}, {n, 0, nmax}]
inth = ListInterpolation[T]
Plot3D[inth[s, t], {s, 0, 1}, {t, 0, 1}]
Any help would be mostly welcome!
Marco
I think your "T" is supposed to be a list of 3D points, in which case you should generate it with:
tmax = 1.;
nmax = 10;
deltat = tmax/nmax;
h[t_, s_] := t^2 + s^2;
T = Table[{n*deltat, n*deltat, h[n*deltat, n*deltat]}, {n, 0, nmax}]
inth = ListInterpolation[T]
Plot3D[inth[s, t], {s, 0, 1}, {t, 0, 1}]
Now T[[1]] = {0., 0., 0.} and not {{0.}, {0.}, 0.} as before.

Transform(align) a plane plot into a 3D plot in Mathematica

I have an ODE and I solve it with NDSolve, then I plot the solution on a simplex in 2D.
Valid XHTML http://ompldr.org/vY2c5ag/simplex.jpg
Then I need to transform (align or just plot) this simplex in 3D at coordinates (1,0,0),(0,1,0),(0,0,1), so it looks like this scheme:
Valid XHTML http://ompldr.org/vY2dhMg/simps.png
I use ParametricPlot to do my plot so far. Maybe all I need is ParametricPlot3D, but I don't know how to call it properly.
Here is my code so far:
Remove["Global`*"];
phi[x_, y_] = (1*x*y)/(beta*x + (1 - beta)*y);
betam = 0.5;
betaf = 0.5;
betam = s;
betaf = 0.1;
sigma = 0.25;
beta = 0.3;
i = 1;
Which[i == 1, {betam = 0.40, betaf = 0.60, betam = 0.1,
betaf = 0.1, sigma = 0.25 , tmax = 10} ];
eta[x2_, y2_, p2_] = (betam + betaf + sigma)*p2 - betam*x2 -
betaf*y2 - phi[x2, y2];
syshelp = {x2'[t] == (betam + betaf + sigma)*p2[t] - betam*x2[t] -
phi[x2[t], y2[t]] - eta[x2[t], y2[t], p2[t]]*x2[t],
y2'[t] == (betaf + betam + sigma)*p2[t] - betaf*y2[t] -
phi[x2[t], y2[t]] - eta[x2[t], y2[t], p2[t]]*y2[t],
p2'[t] == -(betam + betaf + sigma)*p2[t] + phi[x2[t], y2[t]] -
eta[x2[t], y2[t], p2[t]]*p2[t]};
initialcond = {x2[0] == a, y2[0] == b, p2[0] == 1 - a - b};
tmax = 50;
solhelp =
Table[
NDSolve[
Join[initialcond, syshelp], {x2, y2, p2} , {t, 0, tmax},
AccuracyGoal -> 10, PrecisionGoal -> 15],
{a, 0.01, 1, 0.15}, {b, 0.01, 1 - a, 0.15}];
functions =
Map[{y2[t] + p2[t]/2, p2[t]*Sqrt[3]/2} /. # &, Flatten[solhelp, 2]];
ParametricPlot[Evaluate[functions], {t, 0, tmax},
PlotRange -> {{0, 1}, {0, 1}}, AspectRatio -> Automatic]
Third day with Mathematica...
You could find a map from the triangle in the 2D plot to the one in 3D using FindGeometricTransformation and use that in ParametricPlot3D to plot your function, e.g.
corners2D = {{0, 0}, {1, 0}, {1/2, 1}};
corners3D = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
fun[pts1_, pts2_] := FindGeometricTransform[Append[pts2, Mean[pts2]],
PadRight[#, 3] & /# Append[pts1, Mean[pts1]],
"Transformation" -> "Affine"][[2]]
ParametricPlot3D[Evaluate[fun[corners2D, corners3D][{##, 0}] & ### functions],
{t, 0, tmax}, PlotRange -> {{0, 1}, {0, 1}, {0, 1}}]
Since your solution has the property that x2[t]+y2[t]+p2[t]==1 it should be enough to plot something like:
functions3D = Map[{x2[t], y2[t], p2[t]} /. # &, Flatten[solhelp, 2]];
ParametricPlot3D[Evaluate[functions3D], {t, 0, tmax},
PlotRange -> {{0, 1}, {0, 1}, {0, 1}}]

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